Question

A body of mass m falls from a height R above the surface of the earth, where R is the radius of the earth. What is the velocity attained by the body on reaching the ground ? Acceleration due to gravity on the surface of the earth is g.

• A. $gR$
• B. $\sqrt{gR}$
• C. $\sqrt{\frac{g}{R}}$
• D. $\frac{g}{R}$

Answer 1

The correct option is B $\sqrt{gR}$

Suppose a particle of mass m is at a distance R from earth's surface. At this point its K.E is 0 and P.E is $\left(\frac{-GMm}{2R}\right)$, where M is the mass of earth.
On the surface of earth, the P.E of the particle will be $\left(\frac{-GMm}{R}\right)$ and it P.E will be say $\left(\frac{1}{2}\right)m{v}^2$.
Now the total energy in these two cases will be same.
$-\frac{GMm}{2R}+0=-\frac{GMm}{R}+\frac{1}{2}m{v}^2$
$\frac{GMm}{2R}=\frac{1}{2}m{v}^2$
$v=\sqrt{\frac{GM}{R}}$
but $g=\frac{GM}{R^2}$
$gR=\frac{GM}{R}$
So, $v=\sqrt{gR}$